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Search: id:A105076
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| A105076 |
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Sequence a(n) such that 60*(a(n)^2)+60*a(n)+1 = j^2 = a square. |
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| 0, 1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267
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OFFSET
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1,3
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COMMENT
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Proof: define a(1)=0, a(2)=1, a(3)=2, a(4)=11 the 4 first values such that 60*(a(n)^2)+60*a(n)+1=j^2= a square then a(n)=8*a(n-2)+3-a(n-4) 60*(a(n)^2)+60*a(n)+1=j^2 ? need to found positive integer solution a(n) for a(n)^2+a(n)-(j^2-1)/60=0 so (2*a(n)+1)^2=(4*(j^2)+56)/60=(j^2+14)/15 so (j^2+14)/15 needs to be = k^2 or j^2=15*(k^2)-14 put k=2*a(n)+1, 15*((2*a(n)+1)^2)-14 = 60*(a(n)^2)+60*a(n)+1. QED
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FORMULA
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a(n) = a(n-1) +8*a(n-2) -8*a(n-3) -a(n-4) +a(n-5). G.f.: x^2*(1+x+x^2)/((1-x)*(x^4-8*x^2+1)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 13 2009]
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CROSSREFS
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Cf. A103200.
Sequence in context: A067931 A067660 A103200 this_sequence A067670 A159879 A017185
Adjacent sequences: A105073 A105074 A105075 this_sequence A105077 A105078 A105079
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KEYWORD
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nonn,new
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AUTHOR
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Pierre CAMI (pierrecami(AT)tele2.fr), Apr 06 2005
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